M.SC. PHYSICS - II YEAR

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1 MANONMANIAM SUNDARANAR UNIVERSITY DIRECTORATE OF DISTANCE & CONTINUING EDUCATION TIRUNELVELI , TAMIL NADU M.SC. PHYSICS - II YEAR DKP26 - NUMERICAL METHODS (From the academic year ) Most Student friendly University - Strive to Study and Learn to Excel For more information visit:

2 1 PAPER 14 NUMERICAL METHODS Unit I Solution of algebraic and transcendental equations Iteration method, bisection method, Newton Raphson method, rate of convergence solution of polynomial equations Brige Vieta method Bairstow method. Unit II Solution of simultaneous equations Direct method Gauss elimination method Gauss-Jordan method iterative methods Gauss seidal iterative method Eigen values and Eigen vectors of matrices Jacobi method for symmetric matrices. Unit III Interpolation Interpolation formula for unequal intervals Lagrange s method Interpolation formula for equal intervals Newton s forward interpolation formula Newton s Backward interpolation formula - least squares approximation method. Unit IV Numerical differentiation and integration Methods based on interpolation Newton s forward difference formula Newton s backward formula numerical integration Quadrature formula (Newton s cote s formula) Trapezoidal rule, Simpson s 1/3 rd rule, 3/8 th rule Gauss quadrature formula Gauss two point formula and three point formula. Unit V Initial value problems Solution of first order differential equations Taylor series method, Euler s method, Runge-Kutta methods (fourth order) Milne s predictor corrector method Adam- Moulton method. Books for study and Reference 1. Numerical methods for scientific and engineering computations Jain and Iyengar. 2. Numerical methods Venkatraman. 3. Numerical methods Sastry. 4. Numerical methods A. Singaravelu.

3 2 Unit I - Solution of algebraic and transcendental equations An equation f(x)=0 which is only a polynomial in x is known as an algebraic equation. eg. X 5 x 4 + x 26 = 0. Whereas, an equation containing transcendental terms, such as exponential terms, trigonometric terms, logarithmic terms etc, are known as transcendental equations. eg. e x x + 5 = o ; x + 5 cos x + 2 = 0 The point at which the function y = f (x) cuts in the x axis is known as the root of the equation (or) zero of the equation. y = f (x) y = f (a) is positive -f(a) y = f (b) is negative b y = f (c) is zero Therefore c is the root a c of the equaton - f(b) Two points a and b are to be identified for the given equation, such that for one of the points a, y=f(a) is positive and for the point b, f(b) is negative. Then the first approximate root is given be x 0 = (a + b) /2.

4 3 Iteration Method : The root of the given equation f(x) = 0 can be determined by iteration method. Let x 0 be the approximate initial root of the given equation calculated as mentioned above. We have to rewrite the given equation in the form of x = Φ (x) The first approximate root is given by x 1 = Φ(x 0 ) The successive approximations are given by x 2 = Φ (x 1 ) x 3 = Φ (x 2 ).. x n = Φ (x n-1 ) The process of finding root is continued until the successive roots are same to the required accuracy. Note The iteration method will give converging results if Φ' (x) < 1. e.g. Consider the equation f (x) = x 2 + x 3 = 0 f (0) =-3, negative; f (1) = -1, negative; f (2) = 3, positive Hence the root lies between 1 and 2. The initial root is calculated as x 0 = (1 + 2 ) / 2 = 1.5 From the given equation, we can write x = 3 x 2 = Φ (x). Differentiating, we get Φ' (x) = - 2 (x)

5 4 Φ' (1.5) = 3 > 1 Therefore this form of Φ(x) will not give converging solutions. If we rewrite the given equation as x ( x + 1 ) = 3; then we can write x = 3 / (x + 1) = Φ (x) Φ' (x) = - 3 / ( 1 + x ) 2 and Φ' (1.5) < 1 Therefore this form of Φ (x) will give converging solutions. Problems: 1. Find the root of the equation x 3 + x 2 1 = 0 by iteration method correct to two decimal places. Given f (x) = x 3 + x 2 1 = 0 f (0) = -1 negative f (1) = 1, positive Therefore the initial root is x 0 = (0 + 1) / 2 = 0.5 We can rewrite the given equation x 3 + x 2 1 = 0 as x 2 ( x + 1 ) = 1 x = 1/ Therefore Φ (x) = 1/ and Φ' (x) = - 1/ 2( x + 1 ) 3/2 Φ (x) x = 0.5 < 1 Therefore this form of Φ(x) will give converging solutions. The first approximate solution is x 1 = Φ ( x 0 ) = 1/ (1+ x 0 ) = 1 / ( ) = x 2 = Φ ( x 1 ) = 1/ ( ) =

6 5 x 3 = Φ ( x 2 ) = 1/ ( ) = x 4 = Φ ( x 3 ) = 1/ ( ) = Since x 3 x 4 upto two decimal places, the root of the given equation is Find the root of the equation 3 x log 10 x 6 = 0 by iteration method. Given f ( x ) = 3 x log 10 x 6 f ( 1 ) = - 3 ( negative ) f ( 2 ) = ( negative ) f ( 3 ) = ( positive ) The root lies between 2 and 3. Let the initial approximate root be, x 0 = 2 We can rewrite the equation 3 x log 10 x 6 = 0 as X = 1/3( 6 + log 10 x ) Therefore Φ (x) = 1/3( 6 + log 10 x ) The first approximate solution is x 1 = Φ( x 0 ) = 1/3( 6 + log 10 x 0 ) = 1/3 ( 6 + log 10 2) = x 2 = Φ( x 1 ) = 1/3 ( 6 + log ) = x 3 = Φ( x 2 ) = 1/3 ( 6 + log ) = Since x 2 x 3 up to three decimal places, the root of the given equation is Find the root of the equation f(x) = 3 x cos x 1 = 0 by iteration method. Given f(x) = 3 x cosx -1 = 0 Now f ( 0 ) = negative and f(1) = positive

7 6 Let us take initial root as x 0 = 0.6 We can write x = 1/3 ( 1 + cos x ) = Φ (x) x 1 = Φ ( x 0 ) = 1/3 ( 1 + cos 0.6 ) = x 2 = Φ ( x 1 ) = 1/3 ( 1 + cos ) = x 3 = Φ ( x 2 ) = 1/3 ( 1 + cos ) = x 4 = Φ ( x 3 ) = 1/3 ( 1 + cos ) = Since x 3 x 4 correct to three decimal places, the root of the given equation is Bisection method: If f (a) and f (b) are of opposite signs, then the equation f (x) = 0 will have at least one real root between a and b. The bisection method is useful to find the root between a and b. The first approximate root is taken as the midpoint of the range a and b. i.e. x 0 = (a + b) / 2 Then we have to find f(x 0 ). Let us assume f(a) as positive and f(b) as negative. Let f (x 0 ) be negative. Then the root lies between a and x 0. If f (x 0 ) is positive, then the root lies between x 0 and b. We have to bisect the interval in which the root lies and the process is to be repeated. Y -- f(a) a.x 2. x 1 b -- f(b)

8 7 In the above diagram, for the given curve y = f (x), f (a) is positive and f (b) is negative. The first approximate root is the midpoint of a and b i.e. x 1 = (a + b) / 2 f (x 1 ) is negative. Therefore the next approximation is the midpoint of x 1 and a x 2 = (a + x 1 ) / 2 Similarly x 3 = (x 1 + x 2 )/ 2 The process is repeated until two successive roots are equal to the required degree of approximation. Problems : 1. Find the root of the equation x 3 x 1 = 0 correct to two decimal places by bisection method. Let f (x) = x 3 x 1 f (0) = -1 f (1) = -1 ie negative negative f (1.5) = is positive Therefore the initial root is x 0 = ( ) / 2 = 1.25 f (x 0 ) = f (1.25) = (1.25) = Since f (x 0 ) is negative, the root lies between 1.25 and 1.5. approximate root is The next x 1 = ( )/ 2 = f (1.375) = (1.375) = Since f (1.375) is positive, the next root lies between 1.25 and Ie. x 2 = ( ) / 2 =

9 8 f (1.3125) = (1.3125) = (negative) Therefore the root lies between and x 3 = ( ) / 2 = f (x 3 ) = f (1.3438) = (Positive) Therefore the root lies between and Ie. x 4 = ( ) / 2 = f (x 4 ) = (Positive) Therefore the root lies between and Therefore x 5 = ( ) / 2 = Since x 4 x 5 correct to two decimal places, the root is Note: If f(x) 0, then x is the root. Here f (x 5 ) = which is very small and near to zero. Therefore the root is Find the root of the equation x log 10 x 1.2 = 0 by bisection method. Let f (x) = x log 10 x 1.2 f (2) = 2 log = (negative) f (3) = 3 log = (positive) Therefore the root lies between 2 and 3 x 0 =( 2 + 3) / 2 = 2.5 f (x 0 ) = f (2.5) = 2.5 log = Since f (2.5) is negative and f (3) is positive the root lies between 2.5 and 3 Therefore the next approximate root is x 1 = ( ) / 2 = 2.75 f (2.75) = 2.75 log = 0.008

10 9 Since f (2.75) is > 0 and f (2.5) is < o the next root lies between 2.75 and 2.5 Therefore x 2 = ( ) / 2 = f (x 2 ) = f (2.625) = Therefore the next approximation is x 3 = ( ) / 2 = f (x 3 ) = f (2.6875) 0 Therefore the root of the equation is Newton- Raphson Method: Let x 0 be the approximate root of the equation f (x) = 0 Let x 1 = x 0 + h be the exact root of the equation Therefore f (x 1 ) = 0 By Taylor s series expansion, we can write f (x 1 ) = f (x 0 + h) = f (x 0 ) + h/1! f'(x 0 ) + (h 2 / 2!) f'' (x 0 ) +.. Since h is very small, h 2 is negligibly small and we can ignore the higher order terms. Therefore we can write f (x 1 ) = f (x 0 ) + h f' (x 0 ) = 0 i.e. h = - f (x 0 ) / f' (x 0 ) Therefore x 1 = x 0 + h = x 0 - f (x 0 ) / f' (x 0 ) The next approximation is x 2 = x 1 - f (x 1 ) / f' (x 1 )

11 10 In general, x n+1 = x n - f (x n ) / f' (x n ) This is known as Newton Raphson iteration formula. Problems: 1. Find the root of the equation x 3 3 x + 1 = 0 by Newton Raphson method. Given f(x) = x 3 3 x + 1 = 0 f (x) = 3 x 2 3 f (1) = = - 1 (Negative) f (2) = = 3 (Positive) Therefore x 0 =(1 + 2) / 2 = 1.5 f (x 0 ) = x = f! (x 0 ) = 3 (1.5) 2 3 = 3.75 x 1 = 1.5 ( /3.75) = f (x 1 ) = (1.5333) 3 3 x = f (x 1 ) = 3 (1.5333) 2 3 = x 2 = (0.0049/4.053) = Since x 2 x 3 the root of the given equation is Find the root of the equation cos x xe x = 0 by Newton Raphson method. f (x) = cos x xe x f ' (x) = - sin x xe x e x Let x 0 = 0.5

12 11 x 1 = x 0 - f (x 0 ) / f! (x 0 ) = 0.5 f (0.5)/ f! (0.5) = 0.5 (0.533/ ) = = x 2 = x 1 - f (x 1 ) / f! (x 1 ) = f (0.5181)/ f! (0.5181) = ( / ) = = x 3 = x 2 - f (x 2 ) / f! (x 2 ) = f (0.5178)/ f! (0.5178) = ( / ) = Since x 2 x 3 the root of the given equation is Rate of convergence of Newton-Raphson method Let α be the root of the equation f(x) = 0 Let x n be the approximate root of the equation and e n be the small error by which x n and α differs Therefore x n = α + e n Similarly x n+1 = α + e n+1 Newton- Raphson formula is x n+1 = x n f(x n ) / f (x n ) α + e n+1 = α + e n - f(α + e n ) / f ( α + e n ) e n+1 = e n [ f(α + e n ) / f ( α + e n ) ] using Taylor s series expansion we can write

13 12 e n+1 = e n - = e n - = e n f! (α) + e 2 n f!! (α ) f (α ) - e n f! (α ) - e 2 n f!! (α )/2 f! (α ) + e n f!! (α) e 2 n f!! (α) = 2! f! (α) [ 1 + e n f!! (α )/ f! (α)] = K e 2 n e n+1 e 2 n i.e. the error in successive steps decreases as the square of error in the previous step. Thus the order of convergence is two. Birge Vieta method : The real root of a polynomial equation f(x) = 0 can be obtained by this method. Let f (x) = a 0 x n + a 1 x n-1 + a 2 x n a n Let ( x r ) be a factor of f (x). f (x) = ( x - r ) q (x) + R where q (x) = b 0 x n-1 + b 1 x n-2 + b 2 x n b n-1 and R is the remainder Let r 0 be the initial approximation to r. Using Newton Raphson method, the close approximation to the root r is given by r 1 = r 0 f (r 0 )/ f! (r 0 ).

14 13 The values of f (r 0 ) and f! (r 0 ) are calculated by synthetic division formula in Brge Vieta method. Substituting f (x) and q (x) in the equation f (x) = ( x - r ) q (x) + R and comparing the coefficients of like power of x on both sides we get b 0 = a 0 b 1 = a 1 + r 0 a 0 b 2 = a 2 + r 0 b R =b n = a n + r 0 b n-1 In the synthetic division method, f (r 0 ) = R = b n and C i = b i + r 0 C i-1 where C 0 = b 0 and f! (r 0 ) = dr / dr 0 = C n-1 Substituting in the formula r 1 = r 0 f (r 0 )/ f! (r 0 ) we get r 1 = r 0 - b n / C n-1 Synthetic division a 0 a 1 a 2.. a n-2 a n-1 an r 0 r 0 b 0 r 0 b 1.. r 0 b n-3 r 0 b n-2 r 0 b n b 0 b 1 b 2.. b n-2 b n-1 bn r 0 r 0 c 0 r 0 c 1.. r 0 c n-3 r 0 c n c 0 c 1 c 2.. c n-2 c n-1.. Using the formula r 1 = r 0 - b n / C n-1 the approximate root r 1 is calculated.

15 14 Replacing the value of r 1 in r 0 in the above synthetic division method the next approximated value r 2 = r 1 - b n / C n-1 can be obtained. Problem: 1. Find the root of the equation x x 3 21 x 2 22 x + 40 = 0 using Birge Vieta method. Perform two iterative. Take initial root as 3.5. Solution: Here a 0 = 1 a 1 = 2 a 2 = -21 a 3 = -22 a 4 = 40 r o = (= b n ) (=c n-1 ) r 1 = r 0 - b n / C n-1 = ( / 76) = Second iteration ( = b n ) (= c n-1 ) r 2 = r 1 - b n / C n-1 = ( / ) = The root of the given equation is

16 15 2. Find the root of the equation x x x - 20 = 0 using Birge Vieta method. Take initial root as 1. Perform two iterations. Solution: Here a 0 = 1 a 1 = 2 a 2 = 10 a 3 = -20 r o = ( = b n ) (= c n-1 ) r 1 = r 0 - b n / C n-1 = 1 + (7 / 17) = The Second iteration is (= b n ) (= c n-1 ) r 2 = r 1 - ( b n / C n-1 ) = ( / ) = The root of the given equation is

17 16 Bairstow Method: In this method, the quadratic factor of the for m x 2 + px + q for an n th degree polynomial f (x) is obtained by iterative process. Let f (x) = a 0 x 4 + a 1 x 3 + a 2 x 2 + a 3 x + a 4 be a fourth degree polynomial. The quadratic factor be x 2 + px + q. The quotient will be of the for m b 0 x 2 + b 1 x + b 2 and the remainder will be Rx + S. Therefore, we can write a 0 x 4 + a 1 x 3 + a 2 x 2 + a 3 x + a 4 = (x 2 + px + q) (b 0 x 2 + b 1 x + b 2 ) + Rx + S. Equating the coefficients of like powers of x on both sides, we can write b 0 = a 0 b 1 = a 1 pb 0 b 2 = a 2 pb 1 qb 0 b 3 = a 3 pb 2 qb 1 b 4 = a 4 pb 3 qb 2 in general b k = a k pb k-1 qb k-2 and we assume that R = b 3 and S = b 4 + pb 3 We can write c k = b k pc k-1 qc k-2 with c 0 = b 0 The approximations to p and q are given by p and q. p = b 3 c 2 b 4 c 1 c 2 2 c 1 (c 3 b 3 ) q = b 4 c 2 b 3 (c 3 b 3 ) c 2 2 c 1 (c 3 b 3 ) The approximated solution is given by p 1 = p + p and q 1 = q + q The iterations are repeated to get still accurate values.

18 17 The synthetic division procedure is given by a 0 a 1 a 2 a 3 a 4 -p -pb 0 -pb 1 -pb 2 -pb 3 -q -qb 0 -qb 1 -qb b 0 b 1 b 2 b 3 b 4 -p -pc 0 -pc 1 -pc 2 -q -qc 0 -qc c 0 c 1 c 2 c 3 Note: b 0 = a 0 b 1 = a 1 pb 0 c 0 = b 0 b 2 = a 2 pb 1 qb 0 c 1 = b 1 pc 0 b 3 = a 3 pb 2 qb 1 c 2 = b 2 pc 1 qc 0 b 4 = a 4 pb 3 qb 2 c 3 = b 3 pc 2 qc 1

19 18 Problems: 1. Find the quadratic factor of the form x 2 + px + q from the polynomial X 4 3x 3 4x 2 2x + 8 = 0 by Bairstow method with initial values p 0 = q 0 = 1.5 Solution: p 0 = q 0 = (=b 3 ) (=b 4 ) -p 0 = q 0 = Therefore (c 1 ) 8.75 (=c 2 ) (=c 3 ) p 0 = b 3 c 2 b 4 c 1 c 2 2 c 1 (c 3 b 3 ) = (2.875 x 8.75) + ( x 6) = (8.75) 2 6 ( ) q 0 = b 4 c 2 b 3 (c 3 b 3 ) c 2 2 c 1 (c 3 b 3 ) = ( x 8.75) + (2.875 x 4.125) = (8.75) 2 6 ( ) p 1 = p 0 + p 0 = and q 1 = q 0 + q 0 =

20 19 Second iteration p 1 = q 1 = (b 3 ) (b 4) -p 1 = q 1 = (c 1 ) (c 2 ) (c 3 ) p 1 = b 3 c 2 b 4 c 1 c 2 2 c 1 (c 3 b 3 ) = q 1 = b 4 c 2 b 3 (c 3 b 3 ) c 2 2 c 1 (c 3 b 3 ) = Therefore p 2 = p 1 + p 1 = = q 2 = q 1 + q 1 = = Therefore the required factor is x x Solve the equation x 4 3x x x + 54 = 0 and find the quadratic factor of the form x 2 + px + q by Bairstow s method. Perform two iterations. Take p 0 = q 0 = 2

21 20 Solution: p 0 = q 0 = (b 3 ) 2 (b 4) -p 0 = q 0 = (c 1 ) 40 (c 2 ) -68 (c 3 ) p 0 = b 3 c 2 b 4 c 1 c 2 2 c 1 (c 3 b 3 ) = (40) 2 7 (66) = q 0 = b 4 c 2 b 3 (c 3 b 3 ) c 2 2 c 1 (c 3 b 3 ) = 80-2x66 (40) 2 7 (66) = Therefore p 1 = p 0 + p 0 = = q 1 = q 0 + q 0 = =

22 21 Second iteration p 1 = q 1 = (b 3 ) (b 4) -p 1 = q 1 = (c 1 ) (c 2 ) (c 3 ) p 1 = b 3 c 2 b 4 c 1 c 2 2 c 1 (c 3 b 3 ) = q 1 = b 4 c 2 b 3 (c 3 b 3 ) c 2 2 c 1 (c 3 b 3 ) = Therefore p 2 = p 1 + p 1 = q 2 = q 1 + q 1 = Therefore the required factor is x x ********************************

23 22 UNIT II Solution of Simultaneous Linear Algebraic Equations Gauss elimination Method (Direct Method) This is a direct method based on the elimination of the unknowns by combining equations such that the n equations in n unknowns are reduced to an equivalent upper triangular system which could be solved by back substitution. Consider the n linear equations in n unknowns. a 11 x 1 + a 12 x a 1n x n = b 1 a 21 x 1 + a 22 x a 2n x n = b 2. (1) a n1 x 1 + a n2 x a nn x n = b n where a ij and b i are known constants and x i s are unknowns. The system of equations given in (1) is equivalent to AX = B (2) where a 11 a 12.. a 1n x 1 b 1 a 21 a 22.. a 2n x 2 b 2 A =., X =.. and B = a n1 a n2.. a nn x n b n The augmented coefficient matrix is given by (A,B). a 11 a 12.. a 1n b 1 a 21 a 22.. a 2n b 2 (A, B) =. a n1 a n2.. a nn b n (3)

24 23 One has to reduce the augmented matrix (A, B) given in (3) in to an upper triangular matrix. Considering a 11 as the pivot element, multiply the first row of (3) by ( a i1 /a 11 ) and add to the i th row of (A,B), where i = 2,3, n so that all elements in the first column of (A, B) except a 11 are made zero. Now matrix (A,B) will be of the form a 11 a 12.. a 1n b 1 0 b 22.. b 2n c 2. 0 b n2.. b nn c n (4) Considering b 22 as the pivot, we have to make all elements below b 22 in the second column of (4) as zero. This is achieved by multiplying second row of (4) by b i2 /b 22 and add to the corresponding elements of the i th row (i= 3, 4, n). Now all the elements below b 22 are reduced to zero. Now the augmented matrix in (4) has the elements as given below. a 11 a 12 a 13.. a 1n b 1 0 b 22 b 23.. b 2n c c 33.. c 3n d c n3 c nn k n (5) Continuing the process, all elements below the leading diagonal elements of A are made to zero. Repeating the procedure of making the lower diagonal elements to zero for all the columns, we find that the augmented coefficient matrix (A,B) is converted into an upper triangular matrix as shown below. a 11 a 12 a 13 a 14.. a 1n b 1 0 b 22 b 23 b 24.. b 2n c c 33 c 34.. c 3n d α nn k n (6)

25 24 From (6), the given system of linear equations is equivalent to a 11 x 1 + a 12 x 2 + a 13 x a 1n x n = b 1 b 22 x 2 + b 23 x b 2n x n = c 2 c 33 x c 3n x n = d 3 α nn x n = k n Going from the bottom of these equations, we solve for x n = k n / α nn. Using this in the penultimate equation, we get x n-1 and so on. By this back substitution method, we solve for x n, x n-1, x n-2,.. x 2, x 1 Gauss-Jordan elimination method (Direct Method) In this method, the coefficient matrix A of the system AX= B is converted into a diagonal matrix by making all the off diagonal elements to zero by similarity transformations. Now the system (A,B) will be reduced to the form a b 1 0 b c 2 From (7) we get 0 0 c d α nn k n (7) x n = k n / α nn,.. x 2 = c 2 / b 22, x 1 = b 1 / a 11 Problems 1. Solve the system of equations by (i) Gauss elimination method and by (ii) Gauss-Jordan method. X+2y+z=3, 2x+3y+3z=10, 3x-y+2z=13 Gauss elimination method:

26 25 The set of equations are given in matrix form as x y = Z 13 A X = B The augmented coefficient matrix (A,B) for the given system of equations is given as (A,B) = By making the transformations given by the side of the corresponding row of the matrix (A,B), we get (A,B) = R 2 =R 2 + (-2)R R 3 =R 3 + (-3)R 1 Now take b 22 = -1 as the pivot and make b 32 as zero (A,B) = R 3 = R 3 + (-7) R 2 From this, we get x+ 2y+ z = 3 -y + z = 4-8z = -24 Solving the above equations by back substitution we get z = 3, y = -1 and x = 2 Gauss Jordan method:

27 26 The upper triangular matrix (A,B ) is (A,B) = The off diagonal elements of (A,B) are made to zero by the following transformations given below R 1 = R R 2 (A,B) = R 3 =R 3 *(1/8) R 1 = R R 3 = R 2 = R 2 + R Therefore we get, x = 2, y = -I, and z= 3. By substituting these values in the given equations, we find that the values satisfy the equations. 2. Solve the system of equations given below by Gauss elimination method. 2x + 3y z = 5, 4x + 4y - 3z = 3 and 2x - 3y + 2z = 2 The set of equations are given in matrix form as x y = Z 2 A X = B The augmented coefficient matrix (A,B) is (A,B) =

28 27 Taking a 11 = 2 as the pivot, reduce all elements below in the first column to zero (A,B) = R 2 =R 2 +(-2)*R R 3 =R 3 +(-1)*R 1 Taking the element -2 in the position (2,2) as pivot, reduce the element below that to zero, we get (A,B) = R 3 =R 3 +(-3)*R 2 Hence 2x + 3y z = 5-2y z = -7 6z = 18 Then by back substitution we get z = 3, y = 2 and x = 1. These values are substituted in the given equations and are found to satisfy them. 3. Solve the system of equations by Gauss elimination method. 10 x 2 y + 3 z = 23 2 x + 10 y 5 z = x 4 y + 10 z = 41 The given system of equations are written in augmented matrix form as

29 28 Using the transformations given by the side, we can write 1-1/5 3/10 23/10 R 1 =R /5 3/10 23/ /5-28/5-188/5 R 2 =R 2 2 R 1, 0-17/5 91/10 341/10 R 3 = R 3 3R 1 1-1/5 3/10 23/ /13-47/13 R 2 = R 2 52/5 0-17/5 91/10 341/10 1-1/5 3/10 23/ /13-47/ /26 567/26 R 3 =R /5 R 2 1-1/5 3/10 23/ /13-47/ R 3 = R 3 189/26 Now the coefficient matrix is upper diagonal and using back substitution we get Z=3 y - 7/13 (z) = -47/13 13 y 7 (3) = y = y = - 26 Y=2

30 29 x 1/5 y + 3/10 z = 23/10 10 x 2y + 3 z = x 2 (-2) + 3 (3) = 23 X=1 The solution is x=1, y=2, z=3 4. Solve the given system of equations x + 3y + 3 z = 16, x + 4y +3z = 18, x + 3y + 4z = 19 by Gauss Jordan method. x + 3y + 3 z = 16 x + 4y +3z = 18 x + 3y + 4z = 19 Write the given system of equations in augmented matrix form as Add multiples of the first row to the other rows to make all the other components in the first column equal to zero R 2 = R 2 R R 3 = R 3 R R 1 = R 1 3 R 2,

31 R 1 = R 1 3 R The coefficient matrix finally reduces to the diagonal form and the matrix equation is given by x y = z 3 We get x = 1, y = 2, z = 3 5. Solve the given system of equations 10 x + y + z = 12 2 x + 10 y + z = 13 x + y + 5 z = 7 by Gauss Jordan method. The given system of equations are written in augmented matrix form as Make the element in the first row and first column as 1by 1 1/10 1/10 12/10 R 1 = R

32 31 Add multiples of the first row to the other rows and make all the other components in the first column equal to zero 1 1/10 1/10 12/ /5 4/5 106/10 R 2 =R 2 2 R 1, 0 9/10 49/10 58/10 R 3 = R 3 R 1 Make the element in the second row and second column as 1 1 1/10 1/10 12/ /49 53/49 R 2 =R 2 49/5 0 9/10 49/10 58/10 Add multiples of the second row to the other rows to make all the other components in the second column equal to zero R 1 = R 1 (1/10 )R /49 53/ R 3 =R 3 (9/10) R 2, Make the element in the third row and third column as /49 53/ R 3 =R Add multiples of the third row to the other rows to make the components in the third column equal to zero R 1 R R 3, R 2 R 2 4/49 R

33 32 The matrix finally reduces to the form given by x y = z 1 Therefore x = 1, y = 1, z = 1 Gauss Seidel Iterative Method Let the given system of equations be a 11 x 1 + a 12 x 2 + a 13 x a 1n x n = C 1 a 21 x 1 + a 22 x 2 + a 23 x a 2n x n = C 2 a 31 x 1 + a 32 x 2 + a 33 x a 3n x n = C 3. a n1 x 1 + a n2 x 2 + a n3 x a nn x n = C n The system of equations is first rewritten in the form x 1 = (1/a 11 ) (C 1 - a 12 x 2 - a 13 x a 1n x n ). (1) x 2 = (1/a 22 ) (C 2 a 21 x 1 - a 23 x a 2n x n ). (2) x 3 = (1/a 33 ) (C 3 a 31 x 1 - a 32 x a 3n x n ). (3).. X n = (1/a nn ) (C n a n1 x 1 - a n2 x a n,n-1 x n-1 ). (4) First let us assume that x 2 = x 3 = x 4 = = x n = 0 in (1) and find x 2. Let it be x 1 *. Putting x 1 * for x 1 and x 3 = x 4 =.. = x n = 0 in (2) we get the value for x 2 and let it be x 2 *. Putting x 1 * for x 1 and x 2 * for x 2 and x 3 = x 4 =.. = x n = 0 in (3) we get the value for x 3 and let it be x 3 *. In this way we can find the first approximate values for x 1, x 2,.. x n by using the relation x 1 * = (1/a 11 ) (C 1 ) x 2 * = (1/a 22 ) (C 2 a 21 x 1 *) x 3 * = (1/a 33 ) (C 3 a 31 x 1 * - a 32 x 2 *).

34 33 X n * = (1/a nn ) (C n a n1 x 1 * - a n2 x 2 * -.. a n,n-1 x n-1 *) Substituting these values of X* in equations (1),(2),(3),.(4) we get x 1 ** = (1/a 11 ) (C 1 - a 12 x 2 * - a 13 x 3 * a 1n x n *) x 2 ** = (1/a 22 ) (C 2 a 21 x 1 ** - a 23 x 3 * -.. a 2n x n *) x 3 ** = (1/a 33 ) (C 3 a 31 x 1 ** - a 32 x 2 ** -.. a 3n x n *).. X n ** = (1/a nn ) (C n a n1 x 1 ** - a n2 x 2 ** -.. a n,n-1 x n-1 **) Repeating this iteration procedure until two successive iterations give same value, one can get the solution for the given set of equations. This method is efficient for diagonally dominant equations. Diagonally dominant matrix: We say a matrix is diagonally dominant if the numerical value of the leading diagonal element (a ii ) in each row is greater than or equal to the sum of the numerical values of the other elements in that row For example the matrix is diagonally dominant But the matrix is not, since in the Second row, the leading diagonal element 2 is less than the sum of the other two elements viz., 5 and 3 in that row. For the Gauss Seidel method to converge quickly, the coefficient matrix must be diagonally dominant. If it is not so, we have to rearrange the equations in such a way that the coefficient matrix is diagonally dominant and then only we can apply Gauss Seidel method. Problem

35 34 1. Solve the system of equations 4x + 2y + z = 14, x + 5y z = 10, x + y + 8z = 20 using Gauss Seidel iteration method. The given system of equations is 4x + 2y + z = 14 x + 5y z = 10 x + y + 8z = 20..(1)...(2)...(3) The coefficient matrix is diagonally dominant. Hence we can apply Gauss- Seidel method. From (1), (2) and (3) we get x = 1/4 ( 14 2y z)...(4) y = 1/5 ( 10 x + z) z = 1/8 ( 20 x y).(5)...(6) First Iteration Let y = 0, z = 0 in (4) we get x = 14/4 = 3.5 Putting x = 3.5, z = 0 in (5) we get y = 1/5 [ ] = 1.3 Putting x = 3.5, y = 1.3 in (6) we get z = 1/8 [ ] = 15.2/8 = 1.9 Therefore in the first iteration we get x=3.5, y=1.3 and z=1.9 Second Iteration Putting y = 1.3, z = 1.9 in (4) we get x = 1/4 [14 2 (1.3) 1.9] = Putting x = 2.375, z = 1.9 in (5) we get y = 1/5 [ ] = Putting x = 2.375, y = in (6) we get z = 1/8 [ ] = In the second iteration we get x=2.375, y=1.905 and z=1.965

36 35 Third Iteration Putting y = 1.905, z = in (4) we get x = 1/4 [14 2 (1.905) 1.965] = Putting x = 2.056, z = in (5) we get y = 1/5 [ ] = Putting x = 2.056, y = in (6) we get z = 1/8 [ ] = In the third iteration we get x=2.056, y= 1.982, z=1.995 Fourth Iteration Putting y = 1.982, z = in (4) we get x = 1/4 [14 2 (1.982) 1.995] = Putting x = z = in (5) we get y = 1/5 [ ] = Putting x = 2.010, y = in (6) we get z = 1/8 [ ] = In the fourth iteration we get x=2.01, y=1.997, z=1.999 Fifth Iteration Putting y = 1.997, z = in (4) we get x = 1/4 [14 2 (1.997) 1.999] = Putting x = z = in (5) we get y = 1/5 [ ] = Putting x = 2.010, y = in (6) we get z = 1/8 [ ] = 2 In the fifth iteration we get x=2, y=2 and z=2 and the values satisfy the equations.

37 36 2. Solve the system of equations x + y + 54 z = 110, 27 x + 6 y z = 85, 6 x + 15 y + 2 z = 72 using Gauss Seidel iteration method. The given system is x + y + 54 z = 110, 27 x + 6 y z = 85, 6 x + 15 y + 2 z = 72 Interchanging the equations 27 x + 6 y z = 85,. (1) 6 x + 15 y + 2 z = 72. (2) x + y + 54 z = 110,. (3) we get a diagonally dominant system of equations. From (1), (2) and (3) we get x = 1/27 ( 85 6 y + z) y = 1/15 ( 2 6 x - 2 z) z = 1/54 ( 110 x y)...(4).(5)...(6) First Iteration Putting y = 0, z = 0 in (4) we get x = 85/27 = Putting x = 3.148, z = 0 in (5) we get y = 1/15 [ ( 0)] = Putting x = 3.148, y = in (6) we get z = 1/54 [ ] = In the first iteration we get x=3.148, y=3.5408, z=1.913 Second Iteration Putting y = , z = in (4) we get x = 1/27 [ 85 6 (3.5408) 1.913] = Putting x = , z = in (5) we get

38 37 y = 1/15 [ ( 1.913)] = Putting x = , y = in (6) we get z = 1/54 [ ] = In the second iteration we get x=2.4322, y=3.572, z= Third Iteration Putting y = 3.572, z = in (4) we get x = 1/27 [ 85 6 (3.572) ] = Putting x = , z = in (5) we get y = 1/15 [ ( )] = Putting x = , y = in (6) we get z = 1/54 [ ] = In the third iteration we get x= , y=3.5729, z= Fourth Iteration Putting y = , z = in (4) we get x = 1/27 [ 85 6 (3.5729) ] = Putting x = , z = in (5) we get y = 1/15 [ ( )] = Putting x = , y = in (6) we get z = 1/54 [ ] = In the fourth iteration we get x=2.425, y=3.573, z= Since the values of successive iterations are same, the answer is x=2.425, y=3.573, z=

39 38 Jacobi Method If A is a real symmetric matrix and R is an orthogonal matrix then the transformation given by B = R 1 AR = R -1 AR (since R 1 = R -1 ) transforms A into B and this transformation preserves the symmetry and the eigen values of B. That is, B is also symmetric and the eigen values of B are also those of A. If X is the eigen vector corresponding to the eigen value λ of A 1 then the eigen vector of B corresponding to λ is R -1 X. Further if B happens to be in diagonal matrix, the eigen values of B and hence of A are the diagonal elements of B. Rotation Matrix If P( x, y ) is any point in the xy plane and if OP is rotated (O is the origin) in the clockwise direction through an angle θ, then the new points of P ( x, y ) is given by i.e. x Cosθ -Sinθ x y = Sinθ -Cosθ y x x y = R y where R = Cosθ -Sinθ Sinθ -Cosθ Hence R is called a Rotation Matrix in the xy plane. Here R is also a orthogonal matrix since RR = I Let S 1 be a n x n matrix in which the diagonal elements are 1and all other off diagonal elements are zero except a ii = Cosθ, a jj = Cosθ, a ij = - Sinθ, a ji = Sinθ Now S 1 is orthogonal.

40 39 Perform S 1-1 AS 1 = B 1 and this operation annihilates all b ij to zero in B 1. In the next step, take the largest off-diagonal element b kl in B 1 and annihilate to get B 2 = S 2-1 B 1 S 2 Performing series of such rotation by S 1, S 2, S 3,. After K operations, we get B k = S -1 k S -1 k-1 S -1 1 AS 1 S 2 S k = (S 1 S 2. S k ) -1 A (S 1 S 2. S k ) = S -1 AS Where S = S 1 S 2 S k. If B k is diagonal matrix, we get immediately eigen values of B 1 and hence of A. Since an element annihilated by one plane rotation may not necessarily remain zero during subsequent transformations, the operations may continue so that k. As k, B k will approach a diagonal matrix. The minimum number of rotations required to bring matrix A of order n into a diagonal form is n (n-1)/2. Consider a third order matrix A. we will apply the above procedure to get the eigen values for eigen vectors of A. Choose the largest off diagonal element of A and let it be a ij. The matrix will be reduced into a diagonal one if the rotation angle is given by tan2θ = 2a ij / (a ii - a jj ). That is θ = ½ tan -1 (2a ij / (a ii - a jj )). As an example, if a 13 is the largest off diagonal element of matrix A, then θ = ½ tan -1 (2a 13 / (a 11 - a 33 )) and the rotation matrix S 1 is cos θ 0 -sin θ sin θ 0 cos θ And S 1-1 is given by cos θ 0 sin θ sin θ 0 cos θ So that B 1 = S 1-1 A S 1

41 40 Problems: 1. Find the eigen values and eigen vectors of the matrix by Jacobi s method. A = Solution: Here the largest off-diagonal element is a 13 = a 31 = 2 and a 11 = a 33 = 1 Hence take the rotation matrix S 1 = Cosθ 0 -sinθ Sinθ 0 cosθ Now θ = ½ tan -1 (2a 13 / (a 11 - a 33 )) = ½ tan -1 (4 / 0) = ½ tan -1 =(½ )( /2) ie. θ = /4, S 1 = 1/ 2 0-1/ / 2 0 1/ 2 B 1 = S 1-1 AS 1 = 1/ 2 0 1/ / 2 0-1/ / 2 0 1/ / 2 0 1/ 2 = 1/

42 41 = 1/ = Again annihilate the largest off-diagonal element a 12 = a 21 = 2 in B 1 Also a 11 = 3 a 22 = 3 Take S 2 = cosθ -sinθ 0 Sinθ cosθ Now θ = ½ tan -1 (2a 12 / (a 11 - a 22 )) = ½ tan -1 (4 / 0) = ½ tan -1 =(½ )( /2) ie. θ = /4, S 2 = 1/ 2-1/ 2 0 and -1 S 2 = 1/ 2 1/ 2 0 1/ 2 1/ 2 0-1/ 2 1/ B 2 = S 2-1 B 1 S 2 = 1/ 2 1/ / 2-1/ 2 0-1/ 2 1/ / 2 1/

43 42 = 1/ = 1/ = After 2 rotations, A is reduced to the diagonal matrix B 2. Hence the eigen values of A are 5, 1, -1 S = S 1 S 2 = 1/ 2 0-1/ 2 1/ 2-1/ / 2 1/ 2 0 1/ 2 0 1/ = 1/ = 1/2-1/2-1/ 2 1/ 2 1/ 1 0 ½ -1/2 1/ 2 Hence the corresponding vectors are , 2 and

44 43 2. Find the eigen values and eigen vectors of the matrix by Jacobi s me thod. A = Solution: Here the largest off-diagonal element is a 13 = 1 Let us annihilae this element Take S 1 = Cosθ 0 -sinθ Sinθ 0 cosθ tan 2θ = 2a 13 / (a 11 - a 33 ) = 2 / (2-2) = θ = /4, S 1 = 1/ 2 0-1/ / 2 0 1/ 2 B 1 = -1 S 1 AS 1 = 1/ 2 0 1/ / 2 0-1/ / 2 0 1/ / 2 0 1/ 2 = 1/ = 1/

45 44 = Since B 1 is a diagonal matrix the eigen values of A are 3, 2, 1 The corresponding eigen vectors of A are the columns of S 1, namely , 1 and Find the eigen values and eigen vectors of the given matrix by Jacobi s method. Solution: A = The largest off diagonal element is a 23 = 1 and this is to be annihilated. Therefore, the rotation matrix S 1 will have a 22 = a 33 = cos θ, a 23 = -sinθ, a 32 = sinθ a 11 =1, a 12 = a 21 = 0 ie. S 1 = cosθ -sinθ 0 sinθ cosθ tan 2θ = 2a 23 / (a 22 - a 33 ) = -2 / 0 = θ = /4,

46 45 S 1 = / 2-1/ 2 0 1/ 2 1/ 2 B 1 = S 1-1 AS 1 = / 2 1/ / 2-1/ 2 0-1/ 2 1/ / 2 1/ 2 = 1/ = 1/ = Therefore the eigen values of A are 1, 2, 4 The corresponding eigen vectors of A are the columns of S 1, namely , 1 and ***************************

47 46 UNIT III - INTERPOLATION INTERPOLATION WITH UNEQUAL INTERVALS Lagrange s Interpolation Formula for unequal intervals Let f (x 0 ), f (x 1 ), f (x n ) be the values of the function y = f (x) corresponding to the arguments x 0. x 1, x 2,.. x n. The arguments are not equally spaced. Let f (x) be a polynomial in x of degree n. f (x) can be written as f (x) = a 0 ( x x 1 ) ( x x 2 ) ( x x n ) + a 1 ( x x 0 ) ( x x 2 ). ( x x n ) + + a n ( x x 0 ) ( x x 1 ).. ( x x n-1 ). (1) Where a 0, a 1,.. a n are constants. Their values can be obtained by replacing x = x 0 in (1) we get f (x 0 ) = a 0 ( x 0 x 1 ) ( x 0 x 2 ) ( x 0 x n ) i.e. a 0 = f (x 0 ) ( x 0 x 1 ) ( x 0 x 2 ) ( x 0 x n ). (2) Putting x = x 1 in (1) we get f (x 1 ) = a 1 ( x 1 x 0 ) ( x 1 x 2 ) ( x 1 x n ) i.e. a 1 = f (x 1 ) ( x 1 x 0 ) ( x 1 x 2 ) ( x 1 x n ). (3) Similarly. a 2 = f (x 2 ) ( x 2 x 0 ) ( x 2 x 1 ) ( x 2 x n ). (4) a n = f (x n ) ( x n x 0 ) ( x n x 1 ) ( x 0 x n-1 ). (5)

48 47 Substituting (2), (3), (4), (5) in (1) we get f (x) = ( x x 1 ) ( x x 2 ). ( x x n ) f (x 0 ) ( x 0 x 1 ) ( x 0 x 2 ) ( x 0 x n ) + ( x x 0 ) ( x x 2 ). ( x x n ) f (x 1 ) ( x 1 x 0 ) ( x 1 x 2 ) ( x 1 x n ) ( x x 0 ) ( x x 1 ). ( x x n-1 ) f (x n ) ( x n x 0 ) ( x n x 1 ) ( x n x n-1 ) If we denote f (x 0 ), f (x 1 ),.. f (x n ) by y 0, y 1, y n, we get f (x) = ( x x 1 ) ( x x 2 ). ( x x n ) y 0 ( x 0 x 1 ) ( x 0 x 2 ) ( x 0 x n ) + ( x x 0 ) ( x x 2 ). ( x x n ) y 1 ( x 1 x 0 ) ( x 1 x 2 ) ( x 1 x n ) ( x x 0 ) ( x x 1 ). ( x x n-1 ) y n ( x n x 0 ) ( x n x 1 ) ( x n x n-1 ).(6) and this is known as Lagrange s Interpolation formula.

49 48 Problem: 1. Using Lagrange s interpolation formula calculate y(3) from the data given below. x y (x) Here x 0 = 0 x 1 = 1 x 2 = 2 x 3 = 4 x 4 = 5 x 5 = 6 Y 0 = 1 y 1 = 14 y 2 = 15 y 3 = 5 y 4 = 9 y 5 = 19 Lagrange s interpolation formula is y (x) = (x x 1 ) (x x 2 ) (x x 3 ) (x x 4 ) (x x 5 ) y 0 (x 0 x 1 ) (x 0 x 2 ) (x 0 x 3 ) (x 0 x 4 ) (x 0 x 5 ) + (x x 0 ) (x x 2 ) (x x 3 ) (x x 4 ) (x x 5 ) y 1 (x 1 x 0 ) (x 1 x 2 ) (x 1 x 3 ) (x 1 x 4 ) (x 1 x 5 ) + (x x 0 ) (x x 1 ) (x x 3 ) (x x 4 ) (x x 5 ) y 2 (x 2 x 0 ) (x 2 x 1 ) (x 2 x 3 ) (x 2 x 4 ) (x 2 x 5 ) + (x x 0 ) (x x 1 ) (x x 2 ) (x x 4 ) (x x 5 ) y 3 (x 3 x 0 ) (x 3 x 1 ) (x 3 x 2 ) (x 3 x 4 ) (x 3 x 5 ) + (x x 0 ) (x x 1 ) (x x 2 ) (x x 3 ) (x x 5 ) y 4 (x 4 x 0 ) (x 4 x 1 ) (x 4 x 2 ) (x 4 x 3 ) (x 4 x 5 ) + (x x 0 ) (x x 1 ) (x x 2 ) (x x 3 ) (x x 4 ) y 5 (x 5 x 0 ) (x 5 x 1 ) (x 5 x 2 ) (x 5 x 3 ) (x 5 x 4 )

50 49 Substituting the values, we can write y (x) = (x 1) (x 2) (x 4) (x 5) (x 6) X 1 (0 1) (0 2) (0 4) (0 5) (0 6) + (x 0) (x 2) (x 4) (x 5) (x 6) X14 (1 0) (1 2) (1 4) (1 5) (1 6) + (x 0) (x 1) (x 4) (x 5) (x 6) X15 (2 0) (2 1) (2 4) (2 5) (2 6) + (x 0) (x 1) (x 2) (x 5) (x 6) X5 (4 0) (4 1) (4 2) (4 5) (4 6) + (x 0) (x 1) (x 2) (x 4) (x 6) X6 (5 0) (5 1) (5 2) (5 4) (5 6) + (x 0) (x 1) (x 2) (x 4) (x 5) X19 (6 0) (6 1) (6 2) (6 4) (6 5) Substituting x = 3, we get y(3) = using Lagrange s interpolation formula for unequal intervals find the value of y when x = 10 using the values of x and y given below. x y Given: x 0 = 5, x 1 = 6, x 2 = 9, x 3 = 11, y 0 = 12 y 1 = 12 y 2 = 14 y 3 = 16

51 50 Lagrange s interpolation formula is y (x) = (x x 1 ) (x x 2 ) (x x 3 ) y 0 (x 0 x 1 ) (x 0 x 2 ) (x 0 x 3 ) + (x x 0 ) (x x 2 ) (x x 3 ) y 1 (x 1 x 0 ) (x 1 x 2 ) (x 1 x 3 ) + (x x 0 ) (x x 1 ) (x x 3 ) y 2 (x 2 x 0 ) (x 2 x 1 ) (x 2 x 3 ) + (x x 0 ) (x x 1 ) (x x 2 ) y 3 (x 3 x 0 ) (x 3 x 1 ) (x 3 x 2 ) Substituting the values with x=10, we get y(10) = ( 10 6 ) ( 10 9 ) ( ) x (12) ( 5 6 ) ( 5 9 ) ( 5 11 ) + ( 10 5 ) ( 10 9 ) ( ) x (13) ( 6 5 ) ( 6 9 ) ( 6 11 ) + ( 10 5 ) ( 10 6 ) ( ) x (14) ( 9 5 ) ( 9 6 ) ( 9 11 ) + ( 10 5 ) ( 10 6 ) ( 10 9 ) x (16) ( 11 5 ) ( 11 6 ) ( 11 9 )

52 51 = Y(10) = Use Lagrange s interpolation formula to find the form of the function for the Data given below. x f (x) Solution: Here x 0 = 3, x 1 = 2, x 2 = 1, x 3 = -1, y 0 =3, y 1 =12, y 2 = 15, y 3 = -21 f (x) = ( x -2 ) ( x - 1 ) ( x + 1 ) x 3 ( 3-2 ) ( 3-1 ) ( ) + ( x 3 ) ( x 1 ) ( x +1 ) x 12 ( 2 3) ( 2 1 ) ( ) + ( x 3 ) ( x 1 ) ( x +1 ) x 15 ( 1 3 ) ( 1 2 ) ( ) + ( x 3 ) ( x 1 ) ( x +1 ) x (-21) ( -1 3 ) ( -1 3 ) ( -1 1 ) Simplifying, we get f (x) = x 3-9 x x + 6. f(x) is the required form of function for the given data.

53 52 Newton s Forward Interpolation Formula We know that y 0 = y 1 y 0 i.e. y 1 = y 0 + y 0 = (1 + ) y 0 y 1 = y 2 y 1 i.e. y 2 = y 1 + y 1 = (1 + ) y 1 = (1 + ) 2 y 0 y 2 = y 3 y 1 i.e. y 3 = y 2 + y 2 = (1 + ) y 2 = (1 + ) 3 y 0 In general y n = (1 + ) n y 0 Expanding (1 + ) n by using Binomial theorem we have y n = 1 + n + n ( n-1 ) 2 + n ( n-1 ) ( n-2 ) 3 + y 0 2! 3! y n = y 0 + n y 0 + n ( n-1 ) 2 y 0 + n ( n-1 ) ( n-2 ) 3 y 0 + 2! 3! This result is known as Gregory-Newton forward interpolation formula(or) Newton s formula for equal intervals. Problem: 1. The following table gives the population of a town taken during six censuses. Estimate the increase in the population during the period 1946 to Year Population (in thousands) The difference table is givenbelow.

54 53 x y = f(x) y 2 y 3 y 4 y 5 y 1911 (x o ) 12 (y 0 ) 1 ( y 0 ) ( 2 y 0 ) 7-6 ( 3 y 0 ) ( 4 y 0 ) ( y 0 ) Here x 0 = 1911 h = 10 y 0 = 12 Newton s forward interpolation formula is y (x 0 + nh) = y 0 + n y 0 + n ( n-1 ) 2 y 0 + n ( n-1 ) ( n-2 ) 3 y 0 + 2! 3! The population in the year 1946 ie y(1946) is to be calculated. Here x 0 + nh = 1946 i.e., n X 10 = 1946 i.e., n = 3.5 Therefore y (1946) = 12 + (3.5) (1) + (3.5) (3.5-1) x (3.5) (3.5-1) (3.5-2) x ( - 6) 6 + (3.5) (3.5-1) (3.5-2) (3.5-3) x ( 11) 24 + (3.5) (3.5-1) (3.5-2) (3.5-3) ( 3.5 4) x ( -20) 120

55 54 = = Thousands The population in the year 1948 ie y (1948) is calculated as below. Here x 0 + nh = 1948 i.e n.10 = 1948, i.e., n = 3.7 Therefore y (1946) = 12 + (3.7) (1) + (3.7) (3.7-1) x (3.7) (3.7-1) (3.7-2) x ( - 6) 6 + (3.7) (3.7-1) (3.7-2) (3.7-3) x ( 11) 24 + (3.7) (3.7-1) (3.7-2) (3.7-3) ( 3.7 4) x ( -20) 120 = = thousands Increase in the population during the period 1946 to 1948 is = Population in Population in 1946 = = 2.55 thousands. 2. From the following data, find θ at x = 43. x θ Since the vale to be interpolated is at the beginning of the table, we can use Newton s forward interpolation formula.

56 55 The difference table is given below. x y = f(x) y 40 (=x o ) 184 (=y 0 ) 20 (= y 0 ) 2 y 3 y (= 2 y 0 ) 22 0 (= 3 y 0 ) Here x 0 = 40 h = 10 y 0 = 184 By Newton s formula we have y (x 0 + nh) = y 0 + n y 0 + n ( n-1 ) 2 y 0 + n ( n-1 ) ( n-2 ) 3 y 0 + 2! 3! We have to find the value of y at x=43 ie. y(43) i.e., x 0 + nh = 43 i.e., 40 + n.10 = 43 i.e., n = (43 40) /10 = 0.3 Therefore y (43) = (0.3) (20) + (0.3) (0.3-1) x ( 2 ) 2 = =

57 56 Newton Gregory Formula for Backward Interpolation The backward difference formula for any function f (x) is given as f (x) = f (x) f ( x h ), h being interval of data x. If f (a), f( a+h ), f( a+2h).. f(a+nh) are the ( n+1 ) values for the function f (x) corresponding to an independent values of x at equal intervals x = a, a+h, a+2h,. ( a+nh). We can write f(x) in a polynomial of the form f (x) = a 0 + a 1 { x ( a+nh ) } + a 2 { x ( a + nh) } { x - a +( n 1) h} +. + a n { x (a + nh)} { x + a + (n 1) h}. { x ( a+h )} (1) Where the constant a 0, a 1, a 2, a n, are to be determined. Putting x = a + nh, a + n 1 h,.. a in succession in (1) we get f( a + nh ) = a 0 ie. a 0 = f ( a + nh ) f( a + (n -1) h) = a 0 + a 1 (-h) ie. a 1 = f ( a + nh ) - f ( a + (n 1) h) = f ( a + nh) Similarly, a 2 = h 2 f ( a + nh) a n = h n f ( a + nh) 2! h 2 n! h n Substituting these values of a 0, a 1, a 2,. a n in (1) we get f (x) = f ( a + nh) + { x ( a nh)} f ( a + nh) + h + { x ( a + nh)} { x (a + n-1 h)} * 2 f ( a + nh) 2! h 2 + { x ( a + nh)} { x (a + n-1 h)}. { x ( a + h) } * n f ( a + nh). (2) n! h n This is Newton Gregory formula for backward interpolation.

58 57 Putting u = x ( a + nh ) / h or x = a + nh + hu in (2) we get f (x) = f [ a + h ( u+n )] = f( a + nh) + u f( a + nh ) + u ( u + 1) 2 f( a + nh) + 2! + u ( u + 1 ).. ( u + n 1 ) n f( a + nh) n! This is the usual form of Newton-Gregory interpolation formula. Problems: 1. Using Newton s backward interpolation method, fit a polynomial of degree three for the given data. x y Solution The Newton-Gregory backward interpolation formula is y (x 0 + nh) = y 0 + n y 0 + n ( n-1 ) 2 y 0 + n ( n-1 ) ( n-2 ) 3 y 0 + Here x 0 + n h = x x 0 = 6, n=?, h=1 n = x-6 2! 3! x y y 2 y 3 y = 3 y = y y 0 6= x 0 120=y 0

59 58 Y(x)= 120 +(x-6) 60 +(1/2) (x-6)(x-5) 24 + (1/6) (x-6) (x-5) (x-4) 6 Therefore Y(x)= x 3-3x 2 +2x 2. Find the value of y(63) from the data given below. x y Solution The Newton-Gregory backward interpolation formula is y (x 0 + nh) = y 0 + n y 0 + n ( n-1 ) 2 y 0 + n ( n-1 ) ( n-2 ) 3 y 0 + 2! 3! Here, x 0 + n h = x x 0 = 65, h=5, x= 63, n=? n = (63 65)/5 = -2/5 x y y 2 y 3 y 4 y = 3 y = 4 y = y y 0 65= x =y Y(63) = (- )(-6) + ( )(- )(- +1)(2.84) + ( )(- )(- +1) (- +2) (-1.16) 5 2! 5 5 3! ( )(- )(- +1) (- +2) (- +3) (0.68) 4!

60 59 Y(63) = Y(63) = Using Newton s backward interpolation formula evaluate y(9.5) from the data given below. x y Solution y (x 0 + nh) = y 0 + n y 0 + n ( n-1 ) 2 y 0 + n ( n-1 ) ( n-2 ) 3 y 0 + 2! 3! Here, x 0 + n h = x x 0 = 10, h=10, x= 9.5, n=? n = (9.5 10)/10 = x y y 2 y 3 y 4 y = 3 y 0-3= 4 y = y y 0 10= x 0 101=y 0 Y(9.6) = (-0.5)(8) + ( 1 1 )(-0.5)(-0.5+1)(-4) + ( )(-0.5)(-0.5+1) (-0.5+2) (-1) 2! 3! + ( 1 )(-0.5)(-0.5+1) (-0.5+2) (-0.5+3) (-3) 4!

61 60 Y(9.6) = Y(9.6) = The Method of Least Squares: Given n-paired observations (x 1, y 1 ), (x 2, y 2 ),. (x n, y n ) of two variables x and y and we want to determine a function f (x) such that f (x i ) = y i, i = 1, 2, 3,. n we have to fit a straight line of the form y = a + bx and choose the parameters a and b such that the sum of squares of deviations of the fitted value with the given data is least or minimum. This method of fitting a curve is called least squares fitting method. The sum of squares of deviations of the fitted value with the given value is given by s = n 1 2 [ f ( x i ) y i ] is a function of a 1, a 2, a 3,. a n According to the principle of least squares a s may be determined by the requirement that s is least i.e. i s / a 1 = 0, s / a 2 = 0, s / a i = 0, A set of these equations is called the normal equations. The unknowns in y = f (x) are determined using these normal equations. For the linear equation y=a+bx the residuals are given by v i = a + bx i y i, and the sum of residues is given by s = n 1 2 [ a bx i y i ] Differentiating with respect to a and b, we get s / a = 2. n 1 [ a bx i y i ] = 0 and s / b = 2. n 1 x [ a bx y ] = 0 i i i